On a given workday, the rate, in tons per hour, at which unprocessed ore arrives at a refinery is modeled by O(t) = 20 + 10 cos (t^2/25), where t is measured in hours and 0 ≤ t ≤ 10. At the beginning of the workday, t = 0, the refinery has 100 tons of unprocessed ore. During the hours of operation, the refinery processes ore at a constant rate of 25 tons per hour. Determine the time at which the total rate of ore at the refinery is neither increasing nor decreasing? Justify your answer.

gdp is the best phrase!

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

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step-by-step explanation:

nndncnd cnn

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